Let $H$ be a complex Hilbert space and let $(L,D(L))$ be an adjoint operator on $H$. We know that for any $v \in H$ there exists a Borel measure $\mu_v$ on $H$, such that for any Borel bounded function $f:\sigma(L)\to\mathbb C$ we have $$\langle v,f(L)v\rangle=\int_{\sigma(L)}f(x)\mu_v(dx).$$ Suppose that $0\in \sigma(L)$. I would like to understand the interpretation of $\mu_v(\{0\})$:
$\mu_v(\{0\})$ is the squared norm of the projection of $v$ on $\operatorname{Ker}(L)$.
From the formula above we know if we set $f=\chi_{\{0\}}$, where $\chi$ is the indicator function. Then we have $$\langle v,\chi_{\{0\}}(L)v\rangle=\mu_v(\{0\}).$$ But how can I understand $\chi_{\{0\}}(L)$? Does it have anything to do with projections? Thank you for any comment!
If $\chi_{\{0\}}$ is the indicator function of the set $\{0\}$, then $$ P=\chi_{\{0\}}(A) $$ is a selfadjoint projection, meaning that $P=P^*=P^2$. This follows directly from the usual functional calculus for self-adjoint operators. Furthermore, if $id(\lambda)=\lambda$, then $id(A)=A$ and $PA=AP=id(A)\chi_{\{0\}}(A)=0$ because $id\cdot\chi_{\{0\}}$ is the zero function.